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The heat liberated on complete combustion of $$7.8 \,g$$ benzene is $$327 \,kJ$$. This heat was measured at constant volume and at $$27^oC$$. Calculate the heat of combustion of benzene at constant pressure.


A
3274 kJmol1
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B
1637 kJmol1
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C
3270 kJmol1
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D
None of the above
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Solution

The correct option is B $$-3274\ kJ mol^{-1}$$
Solution:- (A) $$- 3274 \; {kJ}/{mol}$$
Molecular weight of $${C}_{6}{H}_{6} = 78 \; g$$
Heat liberated on combustion of $$7.8 \; g$$ benzene $$= 327 \; kJ$$

Heat liberated on combustion of $$78 \; g$$ benzene $$= \cfrac{327}{7.8} \times 78 = 3270 \; kJ$$
$$\therefore \Delta{E} = -3270\ kJ$$
Here negative sign indicates that the energy is liberated.
$${{C}_{6}{H}_{6}}_{\left( l \right)} + 7 \cfrac{1}{2} {{O}_{2}}_{\left( g \right)} \longrightarrow 6 {C{O}_{2}}_{\left( g \right)} + 3 {{H}_{2}O}_{\left( l \right)}$$

From the above reaction,
$$\Delta{{n}_{g}} = {n}_{P} - {n}_{R} = 6 - \cfrac{15}{2} = - \cfrac{3}{2}$$
Temperature $$\left( T \right) = 25 ℃ = \left( 25 + 273 \right) K = 298 K \; \left( \text{Given} \right)$$

Now from first law of thermodynamics,
$$\Delta{H} = \Delta{E} + \Delta{{n}_{g}} RT$$
$$\Delta{H} = \Delta{E} + \left( -\cfrac{3}{2} \right) \times 8.314 \times {10}^{-3} \times 300$$
$$\Rightarrow \Delta{H} = - 3270 - 3.74 = - 3273.74 \; {kJ}/{mol} \approx -3274 \; {kJ}/{mol}$$

Hence the heat of reaction at constant pressure will be $$- 3274 \; {kJ}/{mol}$$.

Chemistry

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